[ros-users] First Message Missing

[Ken Conley]

rospy.spin() keeps your program from exiting, e.g.

1. Your node tells master it is a publisher
2. Master tells subscribers that there is a new publisher
3. Subscriber attempts to connect to new publisher
4. Your node exits

If (4) completes before the subscriber successfully negotiates a
connection, there is no exchange of messages. You are just getting
lucky when you see the message in rostopic echo. On a slower machine,
there's a good chance you won't.

latching does not save the message on a server -- all ROS
communication is peer to peer. It saves the message inside the
publisher, and republishes it to anyone that connects afterwards, i.e.

1. Your node publishes a message
2. A new subscriber connects

In the latched case, the subscriber gets a message. In the unlatched
case, it does not.

Whether or not to latch a particular topic is an important design
consideration and is worth understanding when constructing larger
systems.

 - Ken

On Sun, Sep 12, 2010 at 5:17 PM, David Lu!! <davidvlu at gmail.com> wrote:
> Thanks Ken, for the help and the quick even-on-a-Sunday response.
>
> Latching seems to work, although I'm not sure why it's applicable.
> Spinning at the end did not.
>
> What I was doing was 'rostopic echo chatter' then running the code I
> pasted above, with the two lines switched and a spin at the end.
> rostopic still doesn't print out the message, unless I add latch=True.
> I could understand it not working if I started rostopic after the
> code, but saving the message on the server seems irrelevant here.
>
> -David!!
>
> On Sun, Sep 12, 2010 at 5:56 PM, Ken Conley <kwc at willowgarage.com> wrote:
>> Hi David,
>>
>> There are a couple of issues here:
>>
>>  1) your program starts and then exits, which means that only
>> subscribers that are really fast will get the message. Your node needs
>> to stay up if you want to guarantee that others receive its message
>>  2) you probably want to enable 'latching', described below
>>
>> ROS topics are like tuning into a streaming broadcast -- if you tune
>> in 15 minutes late, you don't get to see the stat. Imagine you are
>> streaming data from the camera -- you want to start receiving messages
>> the most recent one available.
>>
>> The fact that you're getting slightly different results is generally
>> initialization latency. If you put the Publisher init before the call
>> to init_node(), then the rospy node is able to send this data
>> immediately when it communicates with the master. If you put the
>> Publisher init call after init_node(), it has to go back to the
>> master, which incurs more latency.
>>
>> There is a different behavior you can enable with ROS called
>> 'latching', where a publisher will store the most recently published
>> message indefinitely and send it to any new subscriber. We often use
>> this behavior for data that can be static or slowly changing, like
>> maps.
>>
>> See:
>> http://www.ros.org/wiki/rospy/Overview/Publishers%20and%20Subscribers#rospy.Publisher_initialization
>>
>> You probably want to add a rospy.spin() or other sort of delay at the
>> end as well.
>>
>> HTH,
>> Ken
>>
>> On Sun, Sep 12, 2010 at 3:34 PM, David Lu!! <davidlu at wustl.edu> wrote:
>>> Hey all,
>>> I'm not quite sure if this is a bug, or if it is, where to file it,
>>> but it took me forever to track down the problem.
>>>
>>> Here's my test program based on rospy tutorials. (I'm using C-Turtle
>>> binaries on 10.04)
>>> #!/usr/bin/python
>>> import roslib; roslib.load_manifest('crazy')
>>> import rospy
>>> from std_msgs.msg import String
>>>
>>> def talker():
>>>    pub = rospy.Publisher('chatter', String)
>>>    rospy.init_node('talker', anonymous=True)
>>>    str = "hello world"
>>>    pub.publish(str)
>>>
>>> if __name__ == '__main__':
>>>    try:
>>>        talker()
>>>    except rospy.ROSInterruptException: pass
>>>
>>> Fairly standard, and it works like you would expect. However, if you
>>> switch the order of 7th and 8th lines, i.e.
>>>    rospy.init_node('talker', anonymous=True)
>>>    pub = rospy.Publisher('chatter', String)
>>> ...it does NOT publish the message.
>>>
>>> However, when I add a sleep(1) command after Publisher line, then it
>>> works fine.
>>>
>>> Is this...
>>> A) A bug
>>> B) A situation where I should just sleep
>>> C) A situation where I should publish the message repeatedly until I'm
>>> sure it's received.
>>>
>>> Thanks,
>>> -David!!
>>> _______________________________________________
>>> ros-users mailing list
>>> ros-users at code.ros.org
>>> https://code.ros.org/mailman/listinfo/ros-users
>>>
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